Question

and attains the
A train starts from railway station
speed of 72km/h in
5 seconds. Find its acceleration
5
the​

Answer 1

Given,

Initial velocity u = 0

Final velocity v = 72 km/hr

Time taken t = 5 min

= 5/60 hr

= 1/12 hr

Acceleration a = ?

Distance travelled s = ?

Calculation,

i) Acceleration:-

We know that, a = (v - u) / t

a = (72 - 0) / (1/12)

a = 72 / (1/12)

a = (72 * 12) / 1

a = 72 * 12

a = 864 km/hr²

Acceleration = 864 km/hr² Ans.

ii) The distance travelled by the train for attaining this velocity:-

s = ut + 1/2at² (Second equation of motion)

s = 0 * (1/12) + 1/2 * 864 * (1/12)²

s = 1 * 432 * (1/144)

s = 432 / 144

s = 3 km

Distance = 3 km Ans.

Given,

Initial velocity u = 0

Final velocity v = 72 km/hr

Time taken t = 5 min

= 5/60 hr

= 1/12 hr

Acceleration a = ?

Distance travelled s = ?

Calculation,

i) Acceleration:-

We know that, a = (v - u) / t

a = (72 - 0) / (1/12)

a = 72 / (1/12)

a = (72 * 12) / 1

a = 72 * 12

a = 864 km/hr²

Acceleration = 864 km/hr² Ans.

ii) The distance travelled by the train for attaining this velocity:-

s = ut + 1/2at² (Second equation of motion)

s = 0 * (1/12) + 1/2 * 864 * (1/12)²

s = 1 * 432 * (1/144)

s = 432 / 144

s = 3 km

Distance = 3 km Ans.

Expert Verified Answer